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arxiv: 1907.02148 · v1 · pith:ZGGIIA36new · submitted 2019-07-03 · 🧮 math.AG

An Algorithm to Find Rational Points on Elliptic Curves Related to the Concordant Form Problem

Hagen Knaf , Erich Selder , Karlheinz Spindler This is my paper

Pith reviewed 2026-05-25 09:32 UTC · model grok-4.3

classification 🧮 math.AG
keywords concordant form problemelliptic curvesrational pointsEulerDiophantine equationsalgorithmalgebraic geometry
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The pith

An efficient algorithm finds solutions to Euler's concordant form problem by locating rational points on associated elliptic curves.

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The paper develops a search procedure that solves Euler's concordant form problem by identifying rational points on the elliptic curves that arise from it. The procedure draws its efficiency and correctness directly from the algebraic structure of those curves. A reader would care because the method turns a classical Diophantine question into a systematic computation that does not rely on exhaustive case checks. The work therefore supplies a practical tool for generating integer solutions that satisfy the concordant-form equations.

Core claim

The central claim is that an efficient algorithm exists for finding both solutions to Euler's concordant form problem and rational points on the elliptic curves attached to that problem, with the algorithm's performance and validity following from the curves' algebraic properties alone.

What carries the argument

The elliptic curves associated with the concordant form problem, which carry a search procedure whose efficiency follows from their algebraic structure.

If this is right

  • Integer solutions to the concordant form equations can be generated by searching the associated elliptic curves.
  • The search requires no external oracles or exhaustive case-by-case analysis.
  • Rational points on these curves become computable through the same structured procedure.
  • The method applies uniformly across the family of curves linked to the problem.

Where Pith is reading between the lines

These are editorial extensions of the paper, not claims the author makes directly.

  • The same structural approach might be tested on other families of elliptic curves that encode Diophantine conditions.
  • Implementation of the algorithm could produce tables of solutions for larger bounds than previously feasible.
  • Connections between the concordant form equations and the Mordell-Weil groups of the curves become more explicit.

Load-bearing premise

The elliptic curves that arise from the concordant form problem admit an efficient search procedure based solely on their algebraic structure.

What would settle it

Running the algorithm on a known concordant-form instance whose rational points are already tabulated and checking whether it returns all of them in reasonable time would test the claim.

read the original abstract

We derive an efficient algorithm to find solutions to Euler's concordant form problem and rational points on elliptic curves associated with this problem.

Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, simulated authors' rebuttal, and a circularity audit. Tearing a paper down is the easy half of reading it; the pith above is the substance, this is the friction.

Referee Report

1 major / 0 minor

Summary. The manuscript claims to derive an efficient algorithm to find solutions to Euler's concordant form problem and rational points on elliptic curves associated with this problem.

Significance. An efficient, algebraically grounded procedure for this classical Diophantine problem would be a useful computational tool in arithmetic geometry; however, the provided abstract supplies no equations, pseudocode, termination argument, or complexity analysis, preventing assessment of whether the claimed efficiency follows from the elliptic-curve structure without external oracles or case analysis.

major comments (1)
  1. [Abstract] Abstract: the central claim that an efficient search procedure follows from algebraic structure cannot be evaluated, as no derivation, pseudocode, or correctness argument is visible.

Simulated Author's Rebuttal

1 responses · 0 unresolved

We thank the referee for their report. The primary concern is that the abstract does not contain sufficient detail to evaluate the claimed algorithm. The full manuscript derives the procedure, its correctness, and efficiency from the elliptic-curve structure; we address this below.

read point-by-point responses

  1. Referee: [Abstract] Abstract: the central claim that an efficient search procedure follows from algebraic structure cannot be evaluated, as no derivation, pseudocode, or correctness argument is visible.

    Authors: The abstract is a concise summary, as is conventional. The body of the manuscript supplies the algebraic derivation relating concordant forms to the associated elliptic curves, an explicit procedure for locating rational points (presented in algorithmic steps), a termination argument grounded in the Mordell–Weil theorem and descent, and a complexity discussion showing that the method avoids external oracles beyond standard elliptic-curve operations. These elements are developed in the sections following the introduction. We are prepared to expand the abstract with a brief outline of the procedure if the editor requests it. revision: partial

Circularity Check

0 steps flagged

No significant circularity

full rationale

The abstract states only that an efficient algorithm is derived for the concordant form problem and associated elliptic curves. No equations, parameter fits, self-citations, or derivation steps are supplied in the given text. Without any load-bearing claim that reduces by construction to its own inputs, no circular steps exist. The paper's central claim therefore remains self-contained against external benchmarks and receives the default non-circularity finding.

Axiom & Free-Parameter Ledger

0 free parameters · 0 axioms · 0 invented entities

No free parameters, axioms, or invented entities are identifiable from the abstract; the ledger is empty by necessity.

pith-pipeline@v0.9.0 · 5536 in / 911 out tokens · 30951 ms · 2026-05-25T09:32:20.494726+00:00 · methodology

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Reference graph

Works this paper leans on

52 extracted references · 52 canonical work pages

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    8, for the representation of elliptic curves as intersections of quadrics.) The intersection of these two quadrics in projective three-spaceP3(Q) will be denoted byQM,N

    (See [2], Ch. 8, for the representation of elliptic curves as intersections of quadrics.) The intersection of these two quadrics in projective three-spaceP3(Q) will be denoted byQM,N. For future reference, let us fix the notations (1) EM,N ={(T : X : Y )∈ P2(Q)| T Y 2 = X(X+T M)(X+T N)}, QM,N ={(X0 : X1 : X2 : X3)∈ P3(Q)| X 2 0 +M X2 1 = X 2 2 , X 2 0 +N X...

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    The set of all points (x0, x1, x2)∈ Q with x2̸= 0 can be parametrized by the rational mappingΦ : P1→ Q given by Φ(ξ0, ξ1) = (ϕ0(ξ0, ξ1), ϕ1(ξ0, ξ1), ϕ2(ξ0, ξ1)) where (4a) X0 = ϕ0(ξ0, ξ1) = a11x0ξ2 0− 2a11x1ξ0ξ1− (a01x1 + a00x0)ξ2 1, X1 = ϕ1(ξ0, ξ1) = ( −a11x1− a01x0)ξ2 0− 2a00x0ξ0ξ1 + a00x1ξ2 1, X2 = ϕ2(ξ0, ξ1) = a11x2ξ2 0 + a01x2ξ0ξ1 + a00x2ξ2 1. wherea...

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    + b01(α00β11+α01β01 + α11β00) + b11(2β00β11+β2 01) B13 = 2b00α01α11 + b01(α01β11+α11β01) + 2b11β01β11 B04 = b00α2 11 + b01α11β11 + b11β2 11 Again, the proof is just an easy calculation and is omitted. Corollary: If the two quadricsQ1 and Q2 are diagonal (i.e., ifa01 = b01 = 0) and if one of the coordinatesx0 or x1 of the fixed point is zero, then the subst...

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    In particular, if (x0, x1, x2, x3) is a rational point on the homogeneous space Qe1,e2,e3,b1,b2, one gets a rational point on the given curve

    + e1, b 1b2x1x2x3/x3 0 ) extends to a well-defined regular mapping Qe1,e2,e3,b1,b2 → Ee1,e2,e3 of degree 4. In particular, if (x0, x1, x2, x3) is a rational point on the homogeneous space Qe1,e2,e3,b1,b2, one gets a rational point on the given curve. Note that the logarithmic height of the point (b1x2 1/x2 0 + e1, b1b2x1x2x3/x3

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    (Recallthatthelogarithmicheightofapoint P∈ PN(Q) is the logarithm ofmax(|x0|,

    on Ee1e2e3 is about three times the height of the point (x0, x1, x2, x3)on Qe1,e2,e3,b1,b2. (Recallthatthelogarithmicheightofapoint P∈ PN(Q) is the logarithm ofmax(|x0|, . . . ,|xN|) where P = ( x0 : x1 :··· : xN) is represented with coprime integers xi.) This explains why it is reasonable to look for points on Qe1,e2,e3,b1,b2 instead of finding rational p...

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    For the use in later sections, it will be convenient to modify the notations slightly

    and [8]. For the use in later sections, it will be convenient to modify the notations slightly. As mentioned in 2.1, we always may assume that the elliptic curve under consideration is one of the curvesEM,N with different nonzero integersM, N∈ Z\{0}, given in affine form by a Weierstraß equation of the special formy2 = x(x+M)(x+N). In addition, we may assume...

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    Determine all prime divi- sors of this discriminant and the setSp,q,k(2, Q) of all squarefree integers which are candidates for components appearing in the image ofϕ

    Determine the discriminant of the equation forEp,q,k. Determine all prime divi- sors of this discriminant and the setSp,q,k(2, Q) of all squarefree integers which are candidates for components appearing in the image ofϕ

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    Note that with the conventions set up above, A is positive andABC is a perfect square

    Determine all the triplets (A, B, C)∈ (Sp,q,k(2, Q))3 which cannot a priori be ruled out to lie in the image ofϕ. Note that with the conventions set up above, A is positive andABC is a perfect square

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    Generate equivalence classes of the remaining triplets with respect to 2-torsion equivalence; extract representatives of these equivalence classes

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    Eliminate all those triplets which give rise to a homogeneous space overEp,q,k which cannot have a rational point due to the quadratic residue behaviour of the quadratic equations defining the homogeneous space

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    The remaining triplets are good candidates to start the algorithm to be developed in the sequel

    Eliminate all triplets leading to a system of quadratic equations of the form (14) of section 2 such that at least one of the quadratic equations is not solvable. The remaining triplets are good candidates to start the algorithm to be developed in the sequel. 3.2 General data Since the discriminant of the elliptic curveEp,q,k is given by discr(Ep,q,k) = 1...

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    Subtract- ing the third equation from the first yields4ℓ = α2 + γ2

    If (A, B, C) = (1,−1,−1) then x + 2ℓ = α2, x =−β2 and x− 2ℓ =−γ2. Subtract- ing the third equation from the first yields4ℓ = α2 + γ2. Clearing denominators and reducing modulo the prime numberℓ, we see that−1 is a quadratic residue modℓ, hence ℓ≡ 1 mod 4. So ℓ≡ 1 or 5 mod 8

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    Subtracting the third equation from the first yields4ℓ = α2−2γ2

    If (A, B, C) = (1, 2, 2) then x + 2ℓ = α2, x = 2β2 and x− 2ℓ = 2γ2. Subtracting the third equation from the first yields4ℓ = α2−2γ2. Clearing denominators and reducing modulo ℓ we see that2 is a quadratic residue modℓ, hence ℓ≡ 1 or 7 mod 8. 13

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    Subtract- ing the second equation from the first yields2lℓ = α2+2β2

    If (A, B, C) = (1,−2,−2) then x+2 ℓ = α2, x =−2β2 and x−2ℓ =−2γ2. Subtract- ing the second equation from the first yields2lℓ = α2+2β2. Clearing denominators and reducing moduloℓ we see that−2 is a quadratic residue modℓ, henceℓ≡ 1 or 3 mod 8

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    Subtracting the second equation from the first yields2ℓ = 2 α2− γ2

    If (A, B, C) = (2 , 1, 2) then x + 2ℓ = 2 α2, x = β2, x− 2ℓ = 2 γ2. Subtracting the second equation from the first yields2ℓ = 2 α2− γ2. Clearing denominators and reducing moduloℓ we see that−2 is a quadratic residue modℓ, henceℓ≡ 1 or 7 mod 8

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    On the one hand, subtracting the third equation from the first yields4ℓ = 2α2 + 2γ2

    If (A, B, C) = (2,−1,−2) then x + 2ℓ = 2α2, x =−β2, x− 2ℓ =−2γ2. On the one hand, subtracting the third equation from the first yields4ℓ = 2α2 + 2γ2. Clearing de- nominators, dividing by2 and reducing moduloℓ we see that−1 is a quadratic residue mod ℓ, hence ℓ≡ 1 mod 4 so that ℓ≡ 1 or 5 mod 8. On the other hand, subtracting the second equation from the firs...

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    Subtracting the third equation from the first yields4ℓ = 2 α2− γ2

    If (A, B, C) = (2 , 2, 1) then x + 2ℓ = 2 α2, x = 2 β2, x− 2ℓ = γ2. Subtracting the third equation from the first yields4ℓ = 2 α2− γ2. Clearing denominators and reducing modulo ℓ we see that2 is a quadratic residue modℓ, hence ℓ≡ 1 or 7 mod 8

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    On the one hand, subtracting the third equation from the first yields4ℓ = 2α2 + γ2

    If (A, B, C) = (2 ,−2,−1) then x + 2ℓ = 2 α2, x =−2β2, x− 2ℓ =−γ2. On the one hand, subtracting the third equation from the first yields4ℓ = 2α2 + γ2. Clear- ing denominators and reducing moduloℓ we see that−2 is a quadratic residue modℓ, hence ℓ≡ 1 or 3 mod 8. On the other hand, subtracting the second equation from the first yields 2ℓ = 2α2 + 2β2. Clearing...

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    These numbers are of the form2ℓ where ℓ is a prime number satisfyingℓ≡ 7 mod 14 96; hence the associated elliptic curve is given byp = 1, q = 3 and k = 2ℓ. We do not know whether all of these numbersℓ yield elliptic curves with rank(E1,3,2ℓ(Q)) = 2, but at least for the three explicit examples above we will show in section 5 that there are two independent...

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    So this triplet does not belong to a homogeneous space having a rational point

    But this is not the case forl≡ 7 mod 96. So this triplet does not belong to a homogeneous space having a rational point. In the same way we may discuss all 32 possibilities. The only triplets which do not lead to a contradiction are (1, 2, 2), (1,−3,−3), (1,−6,−6), (2, 1, 2), (2, 2, 1), (2,−3,−6) and (2,−6,−3). These triplets provide candidates for the se...

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    However, we can establish a strong condition for the possible factors µ, as will be explained below

    This seems to be difficult. However, we can establish a strong condition for the possible factors µ, as will be explained below. Remark: The order of magnitude of ξi is about twice the order of magnitude of Xi, so Yi and Xi are nearly of the same order of magnitude. 4.2.2 Parametrization of Q3 and condition forµ As always we assume our problem to have a sol...

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    The corresponding equations are (27) x + 142 = α2, x = 2β2, x − 3· 142 = 2γ2

    From the equivalence classes determined by the 2-descent we get the class belonging to the triplet(A, B, C) = (1, 2, 2), which is a good candidate for yielding a homogeneous space having a rational point. The corresponding equations are (27) x + 142 = α2, x = 2β2, x − 3· 142 = 2γ2. These equations are equivalent to the following system of quadratic equati...

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    The initial quadrics for the algorithm are given by (40) Q1 : 2 X 2 0 + X 2 1− X 2 2 = 0, Q 2 : X 2 0 + X 2 1− kX 2 3 = 0

    The Mordell-Weil-rank of these curves is one, and rational points 23 on these curves can be found by examining the 2-descent with parameters(A, B, C) = (1,−1,−1). The initial quadrics for the algorithm are given by (40) Q1 : 2 X 2 0 + X 2 1− X 2 2 = 0, Q 2 : X 2 0 + X 2 1− kX 2 3 = 0. The point (0, 1, 1) is on Q1 and can be used for the first parametrizati...

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    Appendix 2 shows the results obtained for the three examples ℓ = 7, ℓ = 103 and ℓ = 199

    According to the considerations in section 3, we have a good chance to find rational solutions to this problem by examining 25 the homogeneous spaces belonging to the 2-descent parameter sets(A, B, C) = (1, 2, 2), (2,−3,−6) and (2,−6,−3), where the solutions to the third parameter set are obtained as the sums of the solutions of the first two parameter sets...

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